MA1311 (Advanced Calculus) Tutorial sheet 2
[October 7 – 8, 2010]
Name: Solutions
1. For the (rational) function f (x) =
3x2 + 15x − 42
, find
2x2 − 9x + 10
lim f (x)
and
x→2
1
lim f
x→0
x
Solution: For limx→2 f (x) we see that the denominator becomes 0 at x = 2. So does the
numerator. Thus there must be a common factor x − 2 which can be cancelled.
Here we do the factoring by long division of polynomials (although you could surely do it
other ways).
3x + 21
x − 2 3x2 + 15x − 42
3x2 − 6x
21x − 42
21x − 42
0
2x −
x − 2 2x2 −
2x2 −
So
f (x) =
5
9x + 10
4x
−5x + 10
−5x + 10
0
3x2 + 15x − 42
3x + 21
(x − 2)(3x + 21)
=
=
2
2x − 9x + 10
(x − 2)(2x − 5)
2x − 5
(for x 6= 2), and
3x + 21
27
=
= −27
x→2 2x − 5
−1
lim f (x) = lim
x→2
For limx→0 f
1
x
we can rewrite
2
3 x1 + 15 x1 − 42
1
f
=
2
x
2 x1 − 9 x1 + 10
2
3 x1 + 15 x1 − 42 x2
= 1 2
1
2 x − 9 x + 10 x2
3 + 15x − 42x2
2 − 9x + 10x2
3
=
2
=
dy
x3 − 5tx2 + cx + 3
for y =
(where t and c are constants).
dx
(x − t)2
Solution: Using the quotient rule first
d
d
3
2
2
3
2
2
(x
−
5tx
+
cx
+
3)
(x
−
t)
−
(x
−
5tx
+
cx
+
3)
(x
−
t)
dy
dx
= dx
dx
((x − t)2 )2
(3x2 − 10tx + c)(x − t)2 − (x3 − 5tx2 + cx + 3)(2(x − t)(1))
=
(x − t)4
d
(using the chain rule for dx
(x − t)2 )
(3x2 − 10tx + c)(x − t)2 − 2(x − t)(x3 − 5tx2 + cx + 3)
=
(x − t)4
(3x2 − 10tx + c)(x − t) − 2(x3 − 5tx2 + cx + 3)
=
(x − t)3
x3 − 3tx2 + 10t2 x − cx − ct − 6
=
(x − t)3
2. Find
(Probably multiplying this out doesn’t make it any much more tidy.)
3. Find the linear approximation to the graph y = 4x3 − 8x2 + 5x − 21 centered at the point
x = 1.
Solution: The formula we want is f (x) ∼
= f (1) + f 0 (1)(x − 1) and we want this for
f (x) = 4x3 − 8x2 + 5x − 21.
We need f (1) = 4 − 8 + 5 − 21 = −20, f 0 (x) = 12x2 − 16x + 5 and f 0 (1) = 1. So we get
4x3 − 8x2 + 5x − 21 ∼
= −20 + 1(x − 1) = −20 + (x − 1)
Richard M. Timoney
2
(for x near 1).